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I've met the following.. Set

$$b_n:=\int_0^{\pi/2} \cos^{2n+1}(x) \mathrm dx.$$ If I'm not mistaken by recursion one finds that $$b_n=\prod_{j=0}^ n \frac{2n+2}{2n+3}.$$

Wolfram alpha says that $$b_n=\frac{\sqrt\pi \Gamma(n+1)}{2\Gamma\left(n+\frac 32\right)}.$$

How to prove this and, moreover $$\lim_{n\to\infty}\sqrt nb_n=\frac{\sqrt\pi}{2}?$$

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Shouldn't it be $b_n=\prod_{j=0}^ {n-1} \frac{2j+2}{2j+3}?$ – draks ... Oct 4 '12 at 12:15
up vote 0 down vote accepted

First your recursion should yield $$\prod_{j=1}^{n} \left(\dfrac{2j}{2j+1}\right)$$

Now note that $$b_n = \prod_{j=1}^{n} \left(\dfrac{2j}{2j+1}\right) = \prod_{j=1}^{n} \left(\dfrac{j}{j+1/2}\right) = \dfrac{\displaystyle \prod_{j=1}^{n} j}{\displaystyle \prod_{j=1}^{n} \left(j+1/2 \right)} = \dfrac{\sqrt{\pi}}2\dfrac{\Gamma(n+1)}{ \Gamma(3/2) \displaystyle \prod_{j=1}^{n} \left(j+1/2 \right)}$$ Recall that $$\Gamma(z+1) = z \Gamma(z) \,\,\,\, \text{ and } \,\,\,\, \Gamma(3/2) = \sqrt{\pi}/2$$ $$ = \dfrac{\sqrt{\pi}}2\dfrac{\Gamma(n+1)}{ \Gamma(n+3/2)}$$ Once you have this you should be able to get the limit by looking at the asymptotics of $\Gamma(z)$. You could either do by say, Stirling's formula (or) make use of the fact that for large enough $z$, $$\dfrac{\Gamma(z+\alpha)}{\Gamma(z)} \sim z^{\alpha}$$ where $\alpha \in \mathbb{R}$.

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There is a little mistake in your typing, $b_{n}=\int_{0}^{\frac{\pi}{2}}\cos^{2n+1}xdx$ but not $\int_{0}^{1}\cos^{2n+1}xdx$.

There are two methods in calculation:

The first method:

$$\begin{align*} b_{n}&=\int_{0}^{\frac{\pi}{2}}\cos^{2n+1}xdx\\ &=\int_{0}^{\frac{\pi}{2}}\cos^{2n}xd\sin x\\ &=2n\int_{0}^{\frac{\pi}{2}}\sin^{2}x\cos^{2n-1}xdx\\ &=2n\int_{0}^{\frac{\pi}{2}}\cos^{2n-1}x-\cos^{2n+1}xdx\\ &=2nb_{n-1}-2nb_{n} \end{align*}$$

so $b_{n}=\frac{2n}{2n+1}b_{n-1}$, or $\frac{b_{n}}{b_{n-1}}=\frac{2n}{2n+1}$, as $b_{0}=1$, we can obtain that:


With the knowledge of $Gamma$ and $Beta$ functions, we can get the representation as Wolfram Alpha gave.

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